MATH 12002 - CALCULUS I 2.7: Related Rates Part 3: More Examples - - PowerPoint PPT Presentation

MATH 12002 - CALCULUS I §2.7: Related Rates Part 3: More Examples

Professor Donald L. White

Department of Mathematical Sciences Kent State University

D.L. White (Kent State University) 1 / 8

Examples

Example 1

At noon, ship A is 100 km west of ship B. Ship A is sailing south at 35 kilometers per hour (km/h) and ship B is sailing north at 25 km/h. How fast is the distance between the ships changing at 4 pm?

Solution

Let A = distance from starting point to ship A, B = distance from starting point to ship B, Z = distance between the ships. Given: dA

dt = 35 km/h and dB dt = 25 km/h.

Want: dZ

dt at 4 pm.

We need to relate A, B, and Z and then take derivatives with respect to time in order to relate dA

dt , dB dt , and dZ dt .

[Continued →]

D.L. White (Kent State University) 2 / 8

Examples

Example 1 Solution [continued]

We have the following situation at a given time after noon:

✪ ✪ ✪ ✪ ✪ ✪ ✪ ✪ ✪ ✪ ✪ ✪ ✪ ✪ ❄ ✻ q q

SHIP A SHIP B A B 100 km Z 100 km

✻ ❄

A + B

By the Pythagorean Theorem, 1002 + (A + B)2 = Z 2. [Continued →]

D.L. White (Kent State University) 3 / 8

Examples

Example 1 Solution [continued]

We now have Z 2 = 1002 + (A + B)2, and taking derivatives with respect to time yields 2Z · dZ dt = 2(A + B) dA dt + dB dt

• .

Hence dZ dt = A + B Z dA dt + dB dt

• .

Finally, at 4 pm, A = (35)(4) = 140 km, B = (25)(4) = 100 km, and Z =

• 1002 + (140 + 100)2 =
• 1002 + 2402 = 260 km,

and so dZ dt = 140 + 100 260 (35 + 25) = 720 13 ≈ 55.4 km/h.

D.L. White (Kent State University) 4 / 8

Examples

Example 2

A plane flying with a constant speed of 210 miles per hour passes over a ground radar station at an altitude of 2 miles and climbs at an angle of 45◦. How fast is the distance from the plane to the radar station increasing 2 minutes later?

Solution

Let p = distance of plane from the point where it passed over the station, z = distance from the plane to the station. Given: dp

dt = 210 mph.

Want: dz

dt after 2 minutes.

We need to relate p and z and then take derivatives with respect to time in order to relate dp

dt and dz dt .

[Continued →]

D.L. White (Kent State University) 5 / 8

Examples

Example 2 Solution [continued]

We have the following situation at a given time:

✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁

• ✒
• r

r

RADAR 2 mi p z 135◦ PLANE

We use the Law of Cosines to relate p and z, and this says z2 = 22 + p2 − 2(2)(p) cos 135◦ = 4 + p2 − 4p · (− √ 2/2) = 4 + p2 + 2 √ 2p. [Continued →]

D.L. White (Kent State University) 6 / 8

Examples

Example 2 Solution [continued]

We have z2 = p2 + 2 √ 2p + 4, and taking derivatives with respect to time, we obtain 2z dz dt = 2pdp dt + 2 √ 2dp dt , and so dz dt = 2p dp

dt + 2

√ 2 dp

dt

2z = p + √ 2 z · dp dt . We know dp

dt = 210 mph, and so we now need to determine the values of p

and z two minutes after the plane passes over the station. [Continued →]

D.L. White (Kent State University) 7 / 8

Examples

Example 2 Solution [continued]

Since the plane is flying at a speed of 210 mph, 2 minutes (or

1 30 hour)

after it passes over the station, p = 210 · 1

30 = 7 miles,

and z =

• 72 + 2

√ 2(7) + 4 =

• 53 + 14

√ 2 miles. Therefore, dz dt = 7 + √ 2

• 53 + 14

√ 2 · 210 ≈ 207 mph.

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